Newton's Method and Fractals 1

Newton's Method and Fractals 1

NEWTON'S METHOD AND FRACTALS AARON BURTON Abstract. In this paper Newton's method is derived, the general speed of convergence of the method is shown to be quadratic, the basins of attraction of Newton's method are described, and finally the method is generalized to the complex plane. 1. Solving the equation f(x) = 0 Given a function f, finding the solutions of the equation f(x) = 0 is one of the oldest mathematical problems. General methods to find the roots of f(x) = 0 when f(x) is a polynomial of degree one or two have been known since 2000 B.C. [3]. For example, to solve for the roots of a quadratic function ax2 + bx + c = 0 we may utilize the quadratic formula: p −b ± b2 − 4ac x = : 2a Methods to solve polynomials of degree three and four were discovered in the 16th century by the mathematicians dal Ferro, Tartaglia, Cardano, and Ferrai [3]. Finally, in 1826 it was discovered from work done by Abel, that there is no general method to solve polynomials of degree five or greater [3]. Since it is not possible to solve all equations of the form f(x) = 0 exactly, an efficient method of approximating solutions is useful. The algorithm discussed in this paper was discovered by Sir Issac Newton, who formulated the result in 1669. Later improved by Joseph Raphson in 1690, the algorithm is presently known as the Newton-Raphson method, or more commonly Newton's method [3]. Newton's method involves choosing an initial guess x0, and then, through an 1 iterative process, finding a sequence of numbers x0, x1, x2, x3, ··· that converge to a solution. Some functions may have several roots. Later we see that the root which Newton's method converges to depends on the initial guess x0. The behavior of Newton's method, or the pattern of which initial guesses lead to which zeros, can be interesting even for polynomials. When generalized to the complex plane, Newton's method leads to beautiful pictures. In this paper, we derive Newton's method, analyze the method's speed of conver- gence, and explore the basins of attraction2. Finally, we extend Newton's method to the complex plane, and through the aid of computer programming view the complex basins of attraction for several polynomials. 1 2 AARON BURTON Figure 1. The geometry of Newton's method 2. Deriving Newton's method As stated in Section 1, Newton's method involves choosing an initial approxi- mation x0, and then, through an iterative process, finding a sequence of numbers x0, x1, x2, x3, ··· that converges to a solution. Recall our goal is to approximate the root of a function f(x), thus once we chose our x0 we hope to find a point x1 (related to x0 in some way) that is a better approximation for the root of the func- tion f(x). Given a single point x0 there are many ways in which we could proceed to find the point x1. In this paper we will use the tangent line at x0. The tangent line provides the best linear approximation to the function f(x) at the point x0, thus we are implicitly assuming that the tangent line will intersect the x-axis near the desired root. This assumption seems to be valid based on Figure 1. In Section 2 we will discuss how this assumption breaks down under certain circumstances. Now, suppose f(x) is a differentiable function on an interval [a; b] for which we wish to approximate the root. We begin by making a guess or estimate of the root's location, that is specifying an initial point (x0; 0). To determine our next estimate (x1; 0), we draw the tangent line through (x0; f(x0)). The point at which the tangent line intersects the x axis is (x1; 0). Using our initial guess x0, the value 0 of the function f(x0) and the slope of the tangent line f (x0) we can then find the equation of the tangent line at (x0; f(x0)) using the point-slope formula: 0 y − f(x0) = f (x0)(x − x0): To solve for the x intercept we set y = 0 and rearrange terms to get 0 −f(x0) = f (x0)(x − x0) 1 Often called the orbit of x0. 2Formally defined in Section 6, a Basin of Attraction is the set of points which converge to a particular root. NEWTON'S METHOD AND FRACTALS 3 −f(x0) x − x0 = 0 f (x0) and finally f(x0) x = x0 − 0 : f (x0) f(x0) The x intercept is our new guess, or estimate, x1. Thus we have, x1 = x0 − 0 . f (x0) To find x2, we begin the whole process over again. However, this time we begin with the point (x1; 0) and solve for the point (x2; 0). Repeating this algorithm generates a sequence of x values x0, x1, x2, ··· by the rule f(xn) 0 xn+1 = xn − 0 ; f (xn) 6= 0 f (xn) we define as the Newton iteration function N(x). Formally, given a differentiable function f, the Newton function for f is: f(x) (1) N(x) = x − : f 0(x) Then x1 = N(x0) 2 x2 = N(x1) = N(N(x0)) = N (x0) and, in general, n xn = N (x0); where the notation N n means N applied n times. 3. Where Newton's method fails One natural question that arises is whether Newton's method will always con- verge to a root. 3.1. Initial guess is a critical point of f(x). Recall from equation (1) that the definition of the Newton iteration function is f(x) N(x) = x − : f 0(x) From this definition we see that N(x) will not exist if f 0(x) = 0. If we chose an initial point where f 0(x) = 0, then Newton's method will fail to converge to a root. 0 Similarly if f (xn) = 0 for some iteration xn, then Newton's method will also fail to converge to a root. The former case is illustrated for f(x) = x3 + 1 in Figure 2. If we happen to choose our initial guess as x = 0, Newton's method fails to converge since the tangent line at x = 0 never intersects the x axis. 3.2. No root to find. Another way in which Newton's method will fail to converge to the root of a function is if there is no root. Consider the graph of f(x) = x2 + 1 in Figure 3. The function f(x) = x2 + 1 never crosses the x axis, and thus there is no possible solution. If we choose an initial guess, it can be proved that Newton's method will chaotically move around the x axis.3 4 AARON BURTON Figure 2. f(x) = x3 + 1. The initial guess coincides with a critical point. 3.3. Periodic cycle. A third way in which Newton's method will fail to converge is if the initial guess or an iteration coincides with a cycle. For example, consider 3 f(x) = x − 2x + 2 and the initial guess of x0 = 1 as shown in Figure 4. With x0 = 1 we see that 13 − 2(1) + 2 x = N(x ) = 1 − = 1 − 1 = 0; 1 0 3(1)2 − 2 and then 03 − 2(0) + 2 x = N(x ) = 0 − = 0 − (−1) = 1: 2 1 3(0)2 − 2 This example is of a cycle with period two, but cycles of other orders may exist as well.4 Often the problems just described can be avoided by choosing our initial point wisely and by looking at the derivatives of the function to be approximated. Usually it is helpful to graph the function f(x) if possible before using Newton's method. 3For further details, refer to Devaney Chapter 13.2 example four. 4See Devaney Chapter 3.3 for an in depth explanation of cycles. NEWTON'S METHOD AND FRACTALS 5 Figure 3. f(x) = x2 + 1. Nonexistent root. Figure 4. f(x) = x3 − 2x + 2. A cycle of period 2. 6 AARON BURTON 4. Convergence A natural extension of Section 3 is the question of convergence. When exactly can we be sure Newton's method will converge to a root? First a few background definitions and a lemma. Definition 4.1. A root r of the equation f(x) = 0 has multiplicity k if f(r) = 0, f 0(r) = 0,··· , f [k−1](r) = 0 but f [k](r) 6= 0. Here f [k](r) is the kth derivative of f. For example, 0 is a root of multiplicity 2 for f(x) = x2 + x3 and of multiplicity 1 for f(x) = x + x3. Definition 4.2. A point x0 is a fixed point of a function f(x) if and only if 0 f(x0) = x0. Moreover, the point x0 is called an attracting fixed point if jf (x0)j < 1. For our purposes it suffices for the reader to note that if a root is an attracting fixed point of the function N(x), then Newton's method will converge to that point. For more information about fixed and attracting fixed points refer Appendix A. Lemma 4.1. If r is a root of multiplicity k for a function f(x), then f(x) may be written in the form f(x) = (x − r)kG(x); where G(r) 6= 0 Proof. Consider the Taylor expansion of a function f(x) centered about the root r. f 00(r) f 000(r) f(x) = f(r) + f 0(r)(x − r) + (x − r)2 + (x − r)3 + ··· 2! 3! Now suppose that the root r has a multiplicity k.

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